A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s². At the end of 2.0 s its angular momentum about the origin is: A) 0 B) (-36 kg. m²/s) C) (+48 kg . m²/s) D) (-96 kg .m²/s) E) (+96 kg . m²/s)

Question

Quizplus · Accepted Answer

The angular momentum about the origin can be calculated as:
L = r x p, where r is the position vector and p is the momentum vector.
At time t = 2.0 s, the block has traveled a distance of (1/2)(11.0 m/s^2)(2.0 s)^2 = 22.0 m on the positive x-axis. Therefore, its position vector is r = (22.0 m)i.
The momentum vector at time t = 2.0 s can be found by using the equation:
p = m(vf - vi), where vf is the final velocity, vi is the initial velocity (which is zero), and m is the mass.
vf = vi + at = (11.0 m/s^2)(2.0 s) = 22.0 m/s
Therefore, p = (2.0 kg)(22.0 m/s)i = (44.0 kg m/s)i.
Taking the cross product of the position and momentum vectors, we get:
r x p = (22.0 m)i x (44.0 kg m/s)i
= (22.0 m)(44.0 kg m/s)k
= 968 kg m^2/s k
The k component represents the magnitude of the angular momentum about the origin, since the position and momentum vectors are perpendicular to each other. The negative sign indicates that the angular momentum is in the negative z direction (i.e. opposite to the positive z axis).
Therefore, the answer is (B) (-36 kg m^2/s k)

A 20-Kg Block Starts from Rest on the Positive X Axis